You have to find the gram formula mass of C6H6 then do mass (g) = mol x GFM
A)
NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺<span>(aq)
- acid </span>a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺
b)
CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)
- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻
Answer:
Explanation:
<u>1) Rate law, at a given temperature:</u>
- Since all the data are obtained at the same temperature, the equilibrium constant is the same.
- Since only reactants A and B participate in the reaction, you assume that the form of the rate law is:
r = K [A]ᵃ [B]ᵇ
<u>2) Use the data from the table</u>
- Since the first and second set of data have the same concentration of the reactant A, you can use them to find the exponent b:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₂ = (1.50)ᵃ (2.50)ᵇ = 2.50 × 10⁻¹ M/s
Divide r₂ by r₁: [ 2.50 / 1.50] ᵇ = 1 ⇒ b = 0
- Use the first and second set of data to find the exponent a:
r₁ = (1.50)ᵃ (1.50)ᵇ = 2.50 × 10⁻¹ M/s
r₃ = (3.00)ᵃ (1.50)ᵇ = 5.00 × 10⁻¹ M/s
Divide r₃ by r₂: [3.00 / 1.50]ᵃ = [5.00 / 2.50]
2ᵃ = 2 ⇒ a = 1
<u>3) Write the rate law</u>
This means, that the rate is independent of reactant B and is of first order respect reactant A.
<u>4) Use any set of data to find K</u>
With the first set of data
- r = K (1.50 M) = 2.50 × 10⁻¹ M/s ⇒ K = 0.250 M/s / 1.50 M = 0.167 s⁻¹
Result: the rate constant is K = 0.167 s⁻¹
Answer:
v = 23.96 cm³
Explanation:
Given data:
Mass = 15.0 g
Density = 0.626 g/cm³
Volume = ?
Solution:
Formula:
D=m/v
D= density
m=mass
V=volume
Now we will put the values in formula:
d = m/v
v = m/d
v = 15 g / 0.626 g/cm³
v = 23.96 cm³
<h3>
Answer:</h3>
1.25 moles (R.T.P.) or 1.34 moles (S.T.P.)
<h3>
Explanation:</h3>
- 1 mole of a gas occupies a volume of 24 liters at room temperature and pressure (R.T.P.)
- On the other hand, 1 mole of a gas will occupy 22.4 Liters at standard temperature and pressure (S.T.P.)
Therefore, at R.T.P.
30.0 Liters will be equivalent to;
= 30.0 L ÷ 24 L
= 1.25 moles
At S.T.P
30.0 Liters will be equivalent to;
= 30.0 L ÷ 22.4 L
= 1.34 moles
Thus, 30.0 L of helium gas are equivalent to 1.25 moles of He at R.T.P. and 1.34 moles at S.T.P.
